Suppose we want to divide the 10 dogs into three groups, one with 3 dogs, one with 5 dogs, and one with 2 dogs.  How many ways can we form the groups such that Fluffy is in the 3-dog group and Nipper is in the 5-dog group?
Place Fluffy in the 3-dog group and Nipper in the 5-dog group.  This leaves 8 dogs remaining to put in the last two spots of Fluffy's group, which can be done in $\binom{8}{2}$ ways.  Then there are 6 dogs remaining for the last 4 spots in Nipper's group, which can be done in $\binom{6}{4}$ ways.  The remaining 2-dog group takes the last 2 dogs.  So the total number of possibilities is $\binom{8}{2} \times \binom{6}{4} = \boxed{420}$.